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Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a continuous function, and $S \subset \mathbb{R}^n$ a bounded set. Now I want to show that $f(S)$ is bounded.

My thoughts: Since $S$ is bounded it's contained in an open ball $B(a,r)$ for some $r>0$. Then I want to show that $f(B(a,r))$ is a bounded set. I know from the continuity of $f$ that given $\varepsilon >0$ there is $\delta >0$ such that $f(B(a,\delta)) \subset B(f(a),\varepsilon)$ for any $a \in \mathbb{R^n}$. So can I just say I could pick the right $\varepsilon$ so that $\delta = r$? But I feel something is amiss here...

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  • $\begingroup$ Hint: show that $\sup\{d(0, f(x)) | x\in S\} < +\infty$ $\endgroup$ – Gribouillis Nov 16 '17 at 5:13
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    $\begingroup$ The closure of $S$ contains $S$ and is also bounded, hence compact. $\endgroup$ – copper.hat Nov 16 '17 at 5:21
  • $\begingroup$ Possible duplicate of Compact operators in finite dimensional spaces $\endgroup$ – Ramanujan Sep 28 '19 at 12:08
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I think compactness is needed here. Find some $M>0$ such that $S\subseteq K:=\{|x|\leq M\}$, $K$ is compact, and so is the image $f(K)$, hence it is closed and bounded, in particular, it is bounded. Now $f(S)\subseteq f(K)$, so $f(S)$ is bounded as well.

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